Quadcopter Altitude Table and Model Choice
A research team tracks a quadcopter's altitude during a short vertical flight. At one-minute intervals starting at t=0, the recorded altitudes are: t=0, h=0; t=1, h=22; t=2, h=40; t=3, h=54; t=4, h=64; t=5, h=70, all in meters. The team notices the altitude gains are getting smaller each minute. One member wants to fit a linear model because the data look roughly straight on a sketch; another wants a quadratic model because the second differences are constant. A third member points out that the quadcopter's battery plan only covers the first five minutes, so predictions beyond t=5 are extrapolations. The team also needs to know when the quadcopter first passes 60 meters, and whether the model predicts a second crossing within a reasonable flight window.
1. Compute the average rate of change on [1,4] and on [4,5]. What do the results imply about a linear model?
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On [1,4], the average rate of change is (h(4) - h(1))/(4 - 1) = (64 - 22)/3 = 42/3 = 14 meters per minute. On [4,5], it is (70 - 64)/(5 - 4) = 6 meters per minute. The rates are not equal, so a single linear model cannot represent the table exactly. The decreasing rates show the altitude gain is slowing, which is consistent with a downward-opening quadratic rather than a straight line.
2. Show that the data are exactly quadratic and find h(t).
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The first differences are 22, 18, 14, 10, 6. The second differences are all -4, which is constant and nonzero. A constant second difference means a quadratic fits exactly. For a quadratic, the second difference equals 2a, so 2a = -4 and a = -2. Write h(t) = -2t^2 + bt + c. Since h(0) = 0, c = 0. Using h(1) = 22 gives -2 + b = 22, so b = 24. Therefore
h(t) = -2t^2 + 24t.
Checking t = 3 gives -18 + 72 = 54, and t = 5 gives -50 + 120 = 70, so the model matches all table values.
3. Use the model to find when the quadcopter reaches 60 meters. Interpret both solutions and assess the extrapolation at t=6.
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Set -2t^2 + 24t = 60. Divide by -2: t^2 - 12t + 30 = 0. By the quadratic formula, t = (12 +/- sqrt(144 - 120))/2 = (12 +/- sqrt(24))/2 = 6 +/- sqrt(6). Thus t is about 3.55 minutes and about 8.45 minutes. The first crossing, near 3.55 minutes, lies inside the recorded interval and is reliable. The second crossing lies beyond t=5 and depends on the quadratic continuing to describe the flight. The vertex is at t=6, where h=72 meters, so the model predicts the quadcopter levels off and then descends. Because the battery plan covers only the first five minutes, the second crossing is an extrapolation and should not be treated as a confirmed flight event.